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%% products.tex
%% 
%% Made by Alex Nelson
%% Login   <alex@tomato>
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%% Started on  Mon Jul 20 12:44:22 2009 Alex Nelson
%% Last update Mon Jul 20 12:44:22 2009 Alex Nelson
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\begin{prob}
We wish to introduce some notion of a ``product'' of two
categories $\ms{A}$ and $\ms{B}$, which has the intuition that
the objects are ``ordered pairs'' of objects from $\ms{A}$ and
$\ms{B}$ respectively, and whose morphisms are ``ordered pairs''
of morphisms from $\ms{A}$ and $\ms{B}$
respectively. Composition, and everything in general, is done
componentwise --- at least, intuitively. We would like to abandon
the use of set theory, and exclusively use category theory instead.
\end{prob}

We'll begin by leading by misexample and use set theoretic terms
to make our intuitions more precise.

Lets first begin by introducing a set theoretic definition of
product categories:
\begin{defn}%\label{defn:}
Given two categories $\ms{A}$ and $\ms{B}$, we defined the
\define{Product Category} $\ms{A}\times\ms{B}$ consisting of
\begin{enumerate}
\item all pairs of objects $(X_{A},Y_{B})$ where
  $X_{A}\in\ob{\ms{A}}$, $Y_{B}\in\ob{\ms{B}}$, and
\item all pairs of objects $(X_{A},Y_{B})$, $(X_{A}',Y_{B}')$ a
  set of morphisms $(f,g)$ where $f:X_{A}\to X_{A}'$, $g:Y_{B}\to Y_{B}'$.
\end{enumerate}
equipped with
\begin{enumerate}
\item for any object $(X_{A},Y_{B})$, an identity morphism
  $(\id{X_{A}},\id{Y_{B}})$;
\item for any pair of morphisms 
$$(T,U)\xrightarrow{\;\;(f,g)\;\;}(T',U')\xrightarrow{\;\;(f',g')\;\;}(T'',U'')$$
we have a morphism
$$(T,U)\xrightarrow{\;\;(f',g')\circ(f,g)\;\;}(T'',U'')$$
which is calculated componentwise $(f',g')\circ(f,g)=(f'\circ
f,g'\circ g)$.
\end{enumerate}
\end{defn}

Now that we have introduced the notion of taking the ``product''
of two categories, we can introduce ``projection functors'' to
recover the components of the product.

\begin{defn}%\label{defn:}
Let $\ms{A},$ $\ms{B}$ be two categories. We define the
\define{Projection Functors}\index{Functor!Projection} $P,Q$ to
be given by
\begin{equation}%\label{eq:}
\ms{A}\xleftarrow{\;\;P\;\;}\ms{A}\times\ms{B}\xrightarrow{\;\;Q\;\;}\ms{B}
\end{equation}
which is specified by
\begin{equation*}%\label{eq:}
P\left((X,Y)\xrightarrow{\;\;(f,g)\;\;}(X',Y')\right) = X\xrightarrow{\;\;f\;\;}X'
\end{equation*}
and
\begin{equation*}%\label{eq:}
Q\left((X,Y)\xrightarrow{\;\;(f,g)\;\;}(X',Y')\right) = Y\xrightarrow{\;\;g\;\;}Y'.
\end{equation*}
\end{defn}

Now we want to generalize this notion of a product to a more
general setting, and in a more category theoretic manner. We have
the product category for $\ms{Set}\times\ms{Set}$ defined as
above, the Cartesian product of elements and morphisms with
everything done componentwise. This is too set theoretic. We
would like to relax the notion of a product a wee bit.

\begin{defn}%\label{defn:}
Let $\ms{C}$ be a category, $X_{1},X_{2}\in\ms{C}$, we define the
\define{product} of $X_1$ and $X_2$ to be an object $P\in\ms{C}$
with two morphisms $\pi_{1}:P\to X_{1}$ and $\pi_{2}:P\to X_{2}$
such that for every object $C\in\ms{C}$ equipped with the
morphisms
\begin{equation}%\label{eq:}
f_{1}:C\to X_{1},\quad\text{and}\quad f_{2}:C\to X_{2}
\end{equation}
we have that there is a unique function $f:C\to P$ such that the
following diagram commute:
\begin{equation}%\label{eq:}
\vcenter{\xymatrix{
&\ar[dl]_{f_{2}}C\ar@{-->}[d]_{f}\ar[dr]^{f_{1}}&\\
X_{2}&\ar[l]^{\pi_{2}}P\ar[r]_{\pi_{1}}&X_{1}
}}
\end{equation}
\noindent Sometimes we denote $P=X_{1}\times X_{2}$ and $f=\<f_{1},f_{2}\>$.
\end{defn}
This may possibly seem like an arbitrary condition to let such a
diagram hold, but it secretly contains two lifting problems
whose solution is $f$.
